Mixing
Are these numbers sufficient for calculating $P(alpha|beta,gamma)$?
$begingroup$
Given three probabilities $Pr(beta|alpha)$, $Pr(gamma|alpha)$ and $Pr(alpha)$; and that $beta$ and $gamma$ are conditionally independent given $alpha$, can $Pr(alpha|beta, gamma)$ be calculated?
My solution:
$$begin{align}
Pr(alpha|beta, gamma) &= frac{Pr(alpha, beta, gamma)}{Pr(beta, gamma)} \
&= frac{Pr(alpha) Pr(beta, gamma|alpha)}{Pr(beta, gamma)}
end{align}$$
Since $beta$ and $gamma$ are independent given $alpha$, we can get:
$Pr(alpha|beta, gamma) = frac{Pr(alpha) Pr(beta|alpha) Pr(gamma|alpha)}{Pr(beta, gamma)}$
Three probabilities in the numerator are given but the denominator is not, then the given constraints are not sufficient to calculate $Pr(alpha|beta, gamma)$.
Am I right?
probability conditional-probability
asked Jan 17 at 14:35
lernerlerner
314217
$endgroup$
add a comment |
$begingroup$
Given three probabilities $Pr(beta|alpha)$, $Pr(gamma|alpha)$ and $Pr(alpha)$; and that $beta$ and $gamma$ are conditionally independent given $alpha$, can $Pr(alpha|beta, gamma)$ be calculated?
My solution:
$$begin{align}
Pr(alpha|beta, gamma) &= frac{Pr(alpha, beta, gamma)}{Pr(beta, gamma)} \
&= frac{Pr(alpha) Pr(beta, gamma|alpha)}{Pr(beta, gamma)}
end{align}$$
Since $beta$ and $gamma$ are independent given $alpha$, we can get:
$Pr(alpha|beta, gamma) = frac{Pr(alpha) Pr(beta|alpha) Pr(gamma|alpha)}{Pr(beta, gamma)}$
Three probabilities in the numerator are given but the denominator is not, then the given constraints are not sufficient to calculate $Pr(alpha|beta, gamma)$.
Am I right?
probability conditional-probability
asked Jan 17 at 14:35
lernerlerner
314217
$endgroup$
add a comment |
$begingroup$
Given three probabilities $Pr(beta|alpha)$, $Pr(gamma|alpha)$ and $Pr(alpha)$; and that $beta$ and $gamma$ are conditionally independent given $alpha$, can $Pr(alpha|beta, gamma)$ be calculated?
My solution:
$$begin{align}
Pr(alpha|beta, gamma) &= frac{Pr(alpha, beta, gamma)}{Pr(beta, gamma)} \
&= frac{Pr(alpha) Pr(beta, gamma|alpha)}{Pr(beta, gamma)}
end{align}$$
Since $beta$ and $gamma$ are independent given $alpha$, we can get:
$Pr(alpha|beta, gamma) = frac{Pr(alpha) Pr(beta|alpha) Pr(gamma|alpha)}{Pr(beta, gamma)}$
Three probabilities in the numerator are given but the denominator is not, then the given constraints are not sufficient to calculate $Pr(alpha|beta, gamma)$.
Am I right?
probability conditional-probability
asked Jan 17 at 14:35
lernerlerner
314217
$endgroup$
Given three probabilities $Pr(beta|alpha)$, $Pr(gamma|alpha)$ and $Pr(alpha)$; and that $beta$ and $gamma$ are conditionally independent given $alpha$, can $Pr(alpha|beta, gamma)$ be calculated?
My solution:
$$begin{align}
Pr(alpha|beta, gamma) &= frac{Pr(alpha, beta, gamma)}{Pr(beta, gamma)} \
&= frac{Pr(alpha) Pr(beta, gamma|alpha)}{Pr(beta, gamma)}
end{align}$$
Since $beta$ and $gamma$ are independent given $alpha$, we can get:
$Pr(alpha|beta, gamma) = frac{Pr(alpha) Pr(beta|alpha) Pr(gamma|alpha)}{Pr(beta, gamma)}$
Three probabilities in the numerator are given but the denominator is not, then the given constraints are not sufficient to calculate $Pr(alpha|beta, gamma)$.
Am I right?
probability conditional-probability
probability conditional-probability
asked Jan 17 at 14:35
lernerlerner
314217
asked Jan 17 at 14:35
lernerlerner
314217
asked Jan 17 at 14:35
lernerlerner
314217
asked Jan 17 at 14:35
lernerlerner
314217
asked Jan 17 at 14:35
lernerlerner
314217
314217
add a comment |
add a comment |
1 Answer
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$begingroup$
You could use Bayes Theorem and then use the independence given $alpha$:
$$mathbb{P}(beta, gamma) = sum_{alpha} mathbb{P}(beta, gamma | alpha)mathbb{P}(alpha)= sum_{alpha}mathbb{P}(beta | alpha) mathbb{P} (gamma | alpha) mathbb{P} (alpha) $$
P.S.: I'm assuming you are given those probabilities for each $alpha$ and $alpha$ spans in a discrete set.
answered Jan 17 at 14:47
HarnakHarnak
1,299512
$endgroup$
$begingroup$
I think the Bayesian tables are given, then $P(beta,gamma)$ becomes calculable.
$endgroup$
– lerner
Jan 17 at 14:57
$begingroup$
Yes, you need either tables or an analytical expression in function of $alpha$.
$endgroup$
– Harnak
Jan 17 at 14:59
1
$begingroup$
No worries. I was about to ask what you meant :)
$endgroup$
– Harnak
Jan 18 at 15:19
$begingroup$
Is it right that we can know the probability for each $alpha$ if $Pr(alpha)$ is given as described in the question?
$endgroup$
– lerner
Jan 19 at 3:32
$begingroup$
Actually, I assumed you were given $mathbb{P}(alpha)$ for every $alpha$. Otherwise you cannot apply Bayes Theorem.
$endgroup$
– Harnak
Jan 21 at 20:46
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
You could use Bayes Theorem and then use the independence given $alpha$:
$$mathbb{P}(beta, gamma) = sum_{alpha} mathbb{P}(beta, gamma | alpha)mathbb{P}(alpha)= sum_{alpha}mathbb{P}(beta | alpha) mathbb{P} (gamma | alpha) mathbb{P} (alpha) $$
P.S.: I'm assuming you are given those probabilities for each $alpha$ and $alpha$ spans in a discrete set.
answered Jan 17 at 14:47
HarnakHarnak
1,299512
$endgroup$
$begingroup$
I think the Bayesian tables are given, then $P(beta,gamma)$ becomes calculable.
$endgroup$
– lerner
Jan 17 at 14:57
$begingroup$
Yes, you need either tables or an analytical expression in function of $alpha$.
$endgroup$
– Harnak
Jan 17 at 14:59
1
$begingroup$
No worries. I was about to ask what you meant :)
$endgroup$
– Harnak
Jan 18 at 15:19
$begingroup$
Is it right that we can know the probability for each $alpha$ if $Pr(alpha)$ is given as described in the question?
$endgroup$
– lerner
Jan 19 at 3:32
$begingroup$
Actually, I assumed you were given $mathbb{P}(alpha)$ for every $alpha$. Otherwise you cannot apply Bayes Theorem.
$endgroup$
– Harnak
Jan 21 at 20:46
add a comment |
$begingroup$
You could use Bayes Theorem and then use the independence given $alpha$:
$$mathbb{P}(beta, gamma) = sum_{alpha} mathbb{P}(beta, gamma | alpha)mathbb{P}(alpha)= sum_{alpha}mathbb{P}(beta | alpha) mathbb{P} (gamma | alpha) mathbb{P} (alpha) $$
P.S.: I'm assuming you are given those probabilities for each $alpha$ and $alpha$ spans in a discrete set.
answered Jan 17 at 14:47
HarnakHarnak
1,299512
$endgroup$
$begingroup$
I think the Bayesian tables are given, then $P(beta,gamma)$ becomes calculable.
$endgroup$
– lerner
Jan 17 at 14:57
$begingroup$
Yes, you need either tables or an analytical expression in function of $alpha$.
$endgroup$
– Harnak
Jan 17 at 14:59
1
$begingroup$
No worries. I was about to ask what you meant :)
$endgroup$
– Harnak
Jan 18 at 15:19
$begingroup$
Is it right that we can know the probability for each $alpha$ if $Pr(alpha)$ is given as described in the question?
$endgroup$
– lerner
Jan 19 at 3:32
$begingroup$
Actually, I assumed you were given $mathbb{P}(alpha)$ for every $alpha$. Otherwise you cannot apply Bayes Theorem.
$endgroup$
– Harnak
Jan 21 at 20:46
add a comment |
$begingroup$
You could use Bayes Theorem and then use the independence given $alpha$:
$$mathbb{P}(beta, gamma) = sum_{alpha} mathbb{P}(beta, gamma | alpha)mathbb{P}(alpha)= sum_{alpha}mathbb{P}(beta | alpha) mathbb{P} (gamma | alpha) mathbb{P} (alpha) $$
P.S.: I'm assuming you are given those probabilities for each $alpha$ and $alpha$ spans in a discrete set.
answered Jan 17 at 14:47
HarnakHarnak
1,299512
$endgroup$
You could use Bayes Theorem and then use the independence given $alpha$:
$$mathbb{P}(beta, gamma) = sum_{alpha} mathbb{P}(beta, gamma | alpha)mathbb{P}(alpha)= sum_{alpha}mathbb{P}(beta | alpha) mathbb{P} (gamma | alpha) mathbb{P} (alpha) $$
P.S.: I'm assuming you are given those probabilities for each $alpha$ and $alpha$ spans in a discrete set.
answered Jan 17 at 14:47
HarnakHarnak
1,299512
edited Jan 17 at 14:50
answered Jan 17 at 14:47
HarnakHarnak
1,299512
answered Jan 17 at 14:47
HarnakHarnak
1,299512
answered Jan 17 at 14:47
HarnakHarnak
1,299512
1,299512
$begingroup$
I think the Bayesian tables are given, then $P(beta,gamma)$ becomes calculable.
$endgroup$
– lerner
Jan 17 at 14:57
$begingroup$
Yes, you need either tables or an analytical expression in function of $alpha$.
$endgroup$
– Harnak
Jan 17 at 14:59
1
$begingroup$
No worries. I was about to ask what you meant :)
$endgroup$
– Harnak
Jan 18 at 15:19
$begingroup$
Is it right that we can know the probability for each $alpha$ if $Pr(alpha)$ is given as described in the question?
$endgroup$
– lerner
Jan 19 at 3:32
$begingroup$
Actually, I assumed you were given $mathbb{P}(alpha)$ for every $alpha$. Otherwise you cannot apply Bayes Theorem.
$endgroup$
– Harnak
Jan 21 at 20:46
add a comment |
$begingroup$
I think the Bayesian tables are given, then $P(beta,gamma)$ becomes calculable.
$endgroup$
– lerner
Jan 17 at 14:57
$begingroup$
Yes, you need either tables or an analytical expression in function of $alpha$.
$endgroup$
– Harnak
Jan 17 at 14:59
1
$begingroup$
No worries. I was about to ask what you meant :)
$endgroup$
– Harnak
Jan 18 at 15:19
$begingroup$
Is it right that we can know the probability for each $alpha$ if $Pr(alpha)$ is given as described in the question?
$endgroup$
– lerner
Jan 19 at 3:32
$begingroup$
Actually, I assumed you were given $mathbb{P}(alpha)$ for every $alpha$. Otherwise you cannot apply Bayes Theorem.
$endgroup$
– Harnak
Jan 21 at 20:46
$begingroup$
I think the Bayesian tables are given, then $P(beta,gamma)$ becomes calculable.
$endgroup$
– lerner
Jan 17 at 14:57
$begingroup$
I think the Bayesian tables are given, then $P(beta,gamma)$ becomes calculable.
$endgroup$
– lerner
Jan 17 at 14:57
$begingroup$
Yes, you need either tables or an analytical expression in function of $alpha$.
$endgroup$
– Harnak
Jan 17 at 14:59
$begingroup$
Yes, you need either tables or an analytical expression in function of $alpha$.
$endgroup$
– Harnak
Jan 17 at 14:59
1
1
$begingroup$
No worries. I was about to ask what you meant :)
$endgroup$
– Harnak
Jan 18 at 15:19
$begingroup$
No worries. I was about to ask what you meant :)
$endgroup$
– Harnak
Jan 18 at 15:19
$begingroup$
Is it right that we can know the probability for each $alpha$ if $Pr(alpha)$ is given as described in the question?
$endgroup$
– lerner
Jan 19 at 3:32
$begingroup$
Is it right that we can know the probability for each $alpha$ if $Pr(alpha)$ is given as described in the question?
$endgroup$
– lerner
Jan 19 at 3:32
$begingroup$
Actually, I assumed you were given $mathbb{P}(alpha)$ for every $alpha$. Otherwise you cannot apply Bayes Theorem.
$endgroup$
– Harnak
Jan 21 at 20:46
$begingroup$
Actually, I assumed you were given $mathbb{P}(alpha)$ for every $alpha$. Otherwise you cannot apply Bayes Theorem.
$endgroup$
– Harnak
Jan 21 at 20:46
add a comment |
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